Divide the following complex numbers. $ \dfrac{2+26i}{5-3i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${5+3i}$ $ \dfrac{2+26i}{5-3i} = \dfrac{2+26i}{5-3i} \cdot \dfrac{{5+3i}}{{5+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(2+26i) \cdot (5+3i)} {(5-3i) \cdot (5+3i)} = \dfrac{(2+26i) \cdot (5+3i)} {5^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(2+26i) \cdot (5+3i)} {(5)^2 - (-3i)^2} = $ $ \dfrac{(2+26i) \cdot (5+3i)} {25 + 9} = $ $ \dfrac{(2+26i) \cdot (5+3i)} {34} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({2+26i}) \cdot ({5+3i})} {34} = $ $ \dfrac{{2} \cdot {5} + {26} \cdot {5 i} + {2} \cdot {3 i} + {26} \cdot {3 i^2}} {34} $ Evaluate each product of two numbers. $ \dfrac{10 + 130i + 6i + 78 i^2} {34} $ Finally, simplify the fraction. $ \dfrac{10 + 130i + 6i - 78} {34} = \dfrac{-68 + 136i} {34} = -2+4i $